Let ${\mathrm{x}}_1,{\mathrm{x}}_2,...,{\mathrm{x}}_{100}$ be in an arithmetic progression, with ${\mathrm{x}}_1=2$ and their mean equal to 200.  If ${\mathrm{y}}_{\mathrm{i}}=\mathrm{i}({\mathrm{x}}_{\mathrm{i}}-\mathrm{i}),1\le \mathrm{i}\le 100$,  then the mean of ${\mathrm{y}}_1,{\mathrm{y}}_2,...,{\mathrm{y}}_{100}$ is

$$\begin{gathered} {\mathrm{x}}_1,{\mathrm{x}}_2,.....{\mathrm{x}}_{100}\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \\ \mathrm{Given} {\mathrm{x}}_1=2 \mathrm{Mean} \mathrm{\mu }=200 \end{gathered}$$
$=\frac{{\displaystyle \sum _{\mathrm{i}=1}^{100}{\mathrm{x}}_{\mathrm{i}}}}{100} \Rightarrow \frac{{\displaystyle \sum _{\mathrm{i}=1}^{100}{\mathrm{x}}_{\mathrm{i}}}}{100}=200 \Rightarrow \sum _{\mathrm{i}=1}^{100}{\mathrm{x}}_{\mathrm{i}}=200\times 100$
$$\begin{gathered} \Rightarrow \frac{100}{2}\left(2{\mathrm{x}}_1+\left(99\right)\mathrm{d}\right)=200\times 100 \Rightarrow 2{\mathrm{x}}_1+99\mathrm{d}=400 \\ \Rightarrow \boxed{\mathrm{d}=4}{\mathrm{x}}_{100}=2+99\mathrm{d}=2+99\times 4=398 \\ \therefore \boxed{{\mathrm{x}}_{100}=398}\Rightarrow {\mathrm{x}}_{\mathrm{i}}=4\mathrm{i}-2 \therefore \mathrm{Mean} \mathrm{of} {\mathrm{y}}_1,{\mathrm{y}}_2,....{\mathrm{y}}_{100} \end{gathered}$$
$=\frac{{\displaystyle \sum _{\mathrm{i}=1}^{100}{\mathrm{y}}_{\mathrm{i}}}}{100}\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}\frac{{\displaystyle \sum _{\mathrm{i}=1}^{100}\mathrm{i}({\mathrm{x}}_{\mathrm{i}}-\mathrm{i})}}{100}=\frac{{\displaystyle \sum _{\mathrm{i}=1}^{100}\mathrm{i}(4\mathrm{i}-2-\mathrm{i}}}{100}=\frac{{\displaystyle \sum _{\mathrm{i}=1}^{100}\mathrm{i}(3\mathrm{i}-2)}}{100}\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}\frac{3{\displaystyle \sum _{\mathrm{i}=1}^{100}{\mathrm{i}}^2-}2{\displaystyle \sum _{\mathrm{i}=1}^{100}\mathrm{i}}}{100}$
$\therefore \mathrm{Mean}=\frac{3\frac{\left(100\right)\left(101\right)\left(201\right)}{6}-\frac{2\times 100\left(101\right)}{2}}{100}=10049.50$